Engineering Mechanics
|
|
- Rodger Mitchell
- 6 years ago
- Views:
Transcription
1 Engineering Mechanics Continued (5) Mohammed Ameen, Ph.D Professor of Civil Engineering
2 B Section Forces in Beams Beams are thin prismatic members that are loaded transversely. Shear Force, Aial Force and Bending Moment Consider the beam shown below: y w() P Consider a cut at a distance of from the left end of the beam. At the cut surface, we have a single force and a single couple as the resultant of the distributed forces.
3 y w() P R A V y M z H Resolve the force into V y and H The couple is M z H is the aial force (AF), V y is the shear force (SF), and M z is the bending moment (BM). The other part of the beam looks like this: w() M z P H V y R B
4 Sign Convention: A force component at a section is positive if the area vector of the crosssection and the force component both have senses either in the positive or in the negative directions of the reference aes. + V y M z H y H M z + V y
5 For 3D beams there are Two shear forces (V y and V z ), Two bending moments (M y and M z ), One aial force (H), and One torsional moment (M y ). y M y V y H V z M z M z
6 Eample: Draw the shear force and bending moment diagrams of the simply supported beam shown below. y 6 kn m 4 m 0 < < m: V y 4 kn; M z 4 kn m H 0 4 kn V M H > > 6 m: V kn; M 4 6 ( ) m 6 kn V M + 1 kn m H 4 kn
7 y 6 kn Loading Diagram m 4 m 4 kn V kn + SFD M 8 kn m + BMD SFD Shear Force Diagram BMD Bending Moment Diagram
8 y W a + b l Wb/l a b Wb/l Wa/l + V SFD M Wab/l + BDM Draw the SFD and the BMD for a s.s. beam subjected to a u.d.l. w kn/m l
9 0 < < l: V wl/ + w wl/ w V M H M wl/. w./ w / (l ) wl/ w kn/m l wl/ wl/ V Loading diagram + wl/ SFD + M wl /8 BMD
10 y 6 kn 10 kn m m m V (kn) M (kn m) < < 6 M V 8.667
11 Wl W W l W Wl V M Eercise: Draw SFD and BMD of: w kn/m l
12 wl / wl w kn/m l wl wl / V M
13 Differential Relations for Equilibrium There eists relations between the loading, the SF and the BM at any section. Consider the equilibrium of an infinitesimal strip of a beam as shown below: y w() d Note that the loading is assumed to be positive if it acts along the positive direction.
14 w()d M V+dV V d a M+dM ΣF y 0: V + (V + dv) + w() d 0 dv d w ΣM a 0: V d M w() d d/ + (M + dm) 0 dm d V
15 w kn/m l Obtain the SF and BM distributions in the s.s beam shown above subjected to a udl. wl/ y w kn/m l wl/ At, w() w; dv d ; V w + A At 0, V wl/; A wl/; V() w( l/) dm d w V w( l/) M() w / wl / + B; At 0, M 0; B 0;
16 wl/ w kn/m l wl/ wl/ V Loading diagram + wl/ SFD + M wl /8 BMD Note: Maimum BM at point of zero SF Maimum SF at point of zero load
17 Eercise 1: A s.s beam is loaded with a distributed load given by w( ) Asin π l Obtain the SF and BM distributions. y l A 0 Eercise : A s.s beam is loaded with a distributed load given by w() w 0 /l. Obtain the SF and BM distributions. y w 0 l
18 Draw the SFD and BMD of the following beams: 8 kn kn/m m m m m 4 kn 8 kn m 1 kn/m m m m m 4 kn 8 knm 1 kn/m 3 m 3 m 3 m 3 m
19 y 5 kn m m 3 m 1 kn V 3 kn m M + kn m
20 8 kn/m m m m m V m M Point of contrafleure
21 6 kn 1 knm 4 kn/m 3 m 3 m 3 m m V (kn) M (knm) 1.5
22 Chains and Cables Relatively fleible chains and cables are often used to support loads Eamples: Suspension bridges Transmission lines Cable is assumed to be perfectly fleible and inetensible Fleibility assumption means that at the centre of any cross-section of the cable only a tensile force is transmitted and no moments can be present this force must be tangential to the cable profile Inetensibility assumptions denotes that the length of the cable remains a constant.
23 A Coplanar Cable with Loading as a Function of (Parabolic cable): Consider a cable suspended between two rigid supports A and B under the action of a loading w() given per unit length measured in the horizontal direction. The cable and the load are assumed to lie in one and the same plane. y l w h
24 Consider an element of the cable as shown below: T + T w T F 0: θ s T cos θ + (T + T) cos (θ + θ) 0 lim 0 or ( T + T ) cos( θ + θ ) T d ( T cos θ ) 0 d T cos θ H a constant, θ + θ cosθ where H is the horizontal component of the cable tension anywhere along the cable A 0
25 ΣF y 0: T sin θ + (T + T) sin (θ + θ) w av 0 or d ( T sinθ ) w d Integrating the above, we get B A, we get That is, T sin θ w () d + C 1 sinθ 1 cosθ H dy 1 tanθ d H w( ) d integrating the above, we obtain B + C y gives the equation of the deflection curve. C 1 and C are found from the boundary conditions. w( ) d + C [ ] w( ) d d + C1 C 1 y + H 1 1
26 Eample: Neglect the self weight of the cable and determine the shape, length and the maimum force in the cable. y w h l We have 1 y w d + C1 + H w + C1 + C H At 0, y 0 C 0 At 0, dy/d 0 C 1 0 Therefore, the deflection curve is y w H C
27 At l/, y h H y wl 8h 4h l h Parabola w H Now, tension in the cable is l 4 T ma corresponds to maimum θ. T H/ cos θ dy w tan θ ; d H θ ma tan 1 wl H T ma H cos tan 1 wl H which can be simplified as T ma wl 1+ l 4h
28 Length of the cable: Alternatively, / 0 1/ / 0 / 0 / l l l l d l h d dy d dy d ds L l h h l l h l 4 sinh / + + Alternatively, This series converges for cables with small slopes (<45 o ); use first few terms only ] [1 4 / 0 4 l h l h l d d dy d dy L l
29 Eample: The left side of a cable is mounted at an elevation of 7 m below the right side. The sag, measured from left support, is 7 m. Find the maimum tension if the cable has uniform loading in the vertical direction of 1500 N/m. 7 m y 7 m 1500 N/m 33 m a Choose the origin as shown above. 1 y w d + C1 + C H At 0, y 0 and y 0; hence, y w H
30 At a, y 7 m; hence 7 wa ; H a 14H w At 33 a, y 14; hence ( 33 a) w 14 ; H And, a m H N T ma occurs at 33 a. T H/cos θ dy w w(33 a) tan θ ; θ d H H T H cosθ ma 35,35.61 N
31 A Coplanar Cable with Loading being the self weight of the cable (Catenary): Now, the loading is a function of s, the position along the cable. We can replace in the earlier equations by s. T + T θ + θ T θ w s d ds ( T cos θ ) 0 T cos θ H a constant A d ds ( T sinθ ) w( s) T sin θ w (s) ds + C 1 B
32 B A, we get dy 1 tanθ d H w( s) ds + C 1 C We cannot integrate the above directly to obtain y. We have, ds d + dy dy ds d ; dy d ds d 1 From [C] we have ds d 1/ 1 1 w( s) ds + C H 1 ds w( s) ds + C d H 1 1/
33 Separating the variables and integrating, we obtain ds + C 1/ w( s) ds + C1 H Eample: Determine the shape of the cable shown below when loaded by self weight only. y s h l Let w be the weight per unit length of the cable. dy d 1 ws w( s) ds + C1 + C a 1 H H
34 With the coordinates as shown in figure, at s 0, dy/d 0. Hence C 1 0. Now, ds + 1/ ws 1 + H C Integrating, we obtain H w ws sinh C H 1 sinh + At 0, s 0. Hence C 0. s H w sinh w H From Eq [a], we get dy d sinh w H
35 H w y cosh + C 3 w H At 0, y 0. Hence C 3 H/w. H w y cosh 1 w H This curve is called a catenary (Latin catena means chain). At l /, y h. h H wl cosh 1 w H This equation can be solved using numerical methods to obtain H. Thus, T ma can be determined.
36 Eample: A large balloon has a buoyant force of 100 lb. It is held by a 150-ft cable whose weight is 0.5 lb/ft. What is the height h of the balloon above the ground when a steady wind causes it to assume the position shown? What is the maimum tension in the cable? y s h 30 o dy d 1 ws + C (a) 1 H
37 At s 0, dy/d tan 30 1/ 3; Hence C 1 1/ 3 ds + 1/ ws H 3 C H w ws H sinh + + C (b) s H w ( C) w sinh H 1 3 (c) From Eq. (a), we have dy ( C w ) sinh d H (d)
38 Integrating the above, we have H ( C ) w y cosh + w H C 3 (e) At s 150, from Eq. (a) we get dy d ws H 3 H 100 tanθ H + V 100 lb 1 3 T (f) V T sin θ 100 lb θ H s h 30 o
39 At 0, s 0, we get from Eq. (b) H w sinh + C C H w 3 sinh From Eq. (f), we can write H 3 H ; (g) 75 1 tan θ θ T 0 ma + ; ma H ma cosθ ma lb From Eq. (g), C sinh 1 3
40 Introduction Friction Forces Friction is the force distribution at the contact surface between two bodies that impedes sliding motion of one body relative to the other. This force distribution is tangential to the contact surface and acts opposite to the direction of any possible movement. Frictional forces are associated with energy dissipation They are at times undesirable, and at times beneficial! Coulomb friction occurs between bodies with dry contact. Lubrication problems are completely different, and come under the purview of fluid mechanics.
41 Major cause of dry friction is associated with microscopic roughness of the surface of contact. Interlocking microscopic protuberances oppose the relative motion between the surfaces. During sliding, some of these protuberances are sheared off and some are melted which eplains the high rate of wear during dry friction. A smooth surface of contact can only support a normal force; a rough one in addition supports a tangential force. P
42 W P N Body on a smooth surface W P Body on a rough surface S N Laws of Coulomb Friction When we slide a block of wood over a surface, we eert continuously increasing load which is completely resisted by friction until the object starts moving, usually with a lurch.
43 P condition of impending motion or impending slippage dynamic static t Idealised plot of applied force P Coulomb in 1781 presented certain conclusions after carrying out eperiments on blocks which tend to move or are moving without rotation. These conclusions are applicable at the condition of impending slippage or once slippage has begun. These are known as Coulomb s laws of friction.
44 The laws of Coulomb friction are: 1. The total force of friction is independent of area of contact.. For low relative velocities between sliding objects, the frictional force is independent of velocity. However, the sliding frictional force is less than the frictional force corresponding to impending slippage. 3. The total frictional force is proportional to the normal force transmitted across the surface of contact. Static Coefficient of Friction Steel on cast iron 0.4 Copper on steel 0.36 Hard steel on hard steel 0.4 Mild steel on mild steel 0.57 Rope on wood 0.7 Wood on wood
45 Thus, the frictional force f is proportional to the normal reaction. That is P W f N f µ N f µ N N Note: Direction of f is such that it opposes the motion. Eample 1: A heavy bo weighing 1 kn rests on a floor. The static coefficient of friction between the surfaces is What is the largest force P, and what is the highest position h for applying this force that will not allow the bo either to slip or to tip? P h 1 kn 1 m
46 Consider the free body diagram shown below corresponding to the condition of impending tipping. ΣF y 0: N 1 kn ΣF 0: P 0.35N 0.35 kn ΣM A 0: 1 kn Ph 1 0.5; h 0.5/ m P h < h ma to avoid tipping. h 1 m N 0.35 N Eample : Will P hold block A in equilibrium, move block A up or is it too small to prevent A from coming down? W A 1000 N W B 400 N P 50 N µ 0. for all contact surfaces. A B 0 o P
47 N A 1000 N 0. N 0 0 f.b.d. 1 N 1 0. N 1 N N 1 B f.b.d. 400 N P 0. N 3 N 3 Find N 1 and N from f.b.d 1. f.b.d corresponds to B moving to left. Find P. If B moves to the right, determine P 1 (reverse directions of P and the frictional forces).
48 N A 1000 N 0. N N 1 N 1 N N 1 B 400 N P 1 0. N 3 N 3
49 Eample: (a) An automobile is shown in the figure below on a roadway inclined at an angle θ with the horizontal. If the coefficient of static and dynamic friction between the tires and the road are 0.6 and 0.5 respectively, what is the maimum inclination θ ma that the car can climb at uniform speed? It has rearwheel drive and has a total loaded weight of 0 kn. The centre of gravity for this loaded condition has been shown in the diagram. cg 400 mm θ 1.4 m 1.7 m
50 We assume that the drive wheels do not spin (i.e., there is no slip between the wheels and the road surface). The maimum friction force possible is µ s times the normal force as shown below: y cg 400 mm µ s N 1 a N 1 θ 0 kn N ΣF 0: 0.6 N 1 0 sin θ ma ΣF y 0: N 1 + N 0 cos θ ma (i) (ii) ΣM a 0: 3.1 N 0 cos θ ma sin θ ma 0.4 (iii)
51 Eliminate N 1 from (i); then eliminate N from (ii); put these in (iii) and simplify to get tan θ ma Hence, θ ma If the drive wheels spin, we need to use µ d in the place of µ s. We will get a smaller value for θ ma in that case (θ ma ). (b) Using the above data, compute the torque needed by the drive wheels to move the car at a uniform speed up an incline with θ 1 0. Wheel dia 450 mm. cg 400 mm f a N kn N
52 In this case, the friction force f has to be determined from Newton s laws (and not by Coulomb s) as there is no impending slippage between the wheel and the road. ΣF 0: f 0 sin 1 0 Hence, f kn. torque Now, looking at the free body diagram of the wheel alone, we get f N 1 torque f r knm (c) What force is needed to tow the car up the incline when it is parked with brakes locked on all the four wheels with θ 1 0? What force is necessary to tow it down the incline?
53 F up cg 0.5 N 1 a N kn 0.5 N N N 1 + N 0 cos 1 0 F up 0.5(N 1 + N ) + 0 sin kn Similarly calculate F down. ( 5.63 kn) cg 0.5 N F 1 0 down a 0 kn 0.5 N 1 N 1 N
54 Eercise: A ladder 4 m long and 00 N weight is placed as shown. A man weighing 600 N climbs up the ladder. At what position will he induce slipping? Coefficient of static friction for all contacts is N B 600 N N B N A N 0. N A N B 0. N A N A + 0.N B 800 N N A N N B N
55 Belt Friction Consider the fleible belt shown in figure wrapped around a portion of a drum. The angle of wrap is β. Assume that the drum is stationary and the belt is just about to move (impending motion). Also, T 1 > T (and hence the belt moves clockwise with respect to the drum). ds θ dθ β T T 1
56 Consider an infinitesimal segment of the belt as a free body diagram. dθ/ dθ/ T t µ dn s dn T + dt n dθ ΣF t 0: dθ dθ T cos + ( T + dt )cos µ sdn 0 or dt µ s dn (i) ΣF n 0: T dθ dθ sin ( T + dt )sin + dn or T dθ dn (ii) 0
57 dt T µ dθ s Integrating the above, we obtain That is T 1 β T dt T 0 T 1 ln µ s β T µ dθ s Or T 1 T e µ s β Thus, the ratio of tensions depend only on the angle of wrap β. Note: (i) β should be in radians. (ii) Appropriate µ should be used.
58 y T 1 W A r T A F If the drum A is forced to the right, as long as the angle of wrap β remains unchanged, the ratio of T 1 /T for impending or actual constant speed slippage will remain unaffected. The torque developed by the belt on the drum as a result of friction is affected by the force F. The torque is obtained as torque T 1 r T r (T 1 T ) r
59 Eample: A drum having a radius of 400 mm as shown in the figure below needs a torque of 500 Nm to get it to start rotating. If µ s 0.5, determine the minimum aial force F needed to create sufficient tension to start the rotation of the drum. T Torque r F T T 1 π µ sβ 180 e e Torque 500 Nm (T 1 T ) r Hence get T 1 and T ; then F (T 1 + T ) cos T
60 Eample: The seaman pulls with 100 N force and wants to stop the motor boat from moving away from the dock. How few wraps n of the rope must he make around the post if the motor boat develops a thrust of 3500 N. (µ s 0. between rope and the post.) T 1 T µ s β e T N; T 100 N Hence β wraps.
61 Eample: What is the maimum weight W that can be supported by the system in the position shown? Pulley B cannot turn. Bar AC is fied to cylinder A, which weighs 500 N. µ s 0.3 for all contact surfaces. B 6 m m W C A
62 6 m T 500 N m 0.3 N 1 C A N 1 ΣF 0: 0.3 N N 1 N 0.3 N ΣF y 0: 0.3 N 1 + N + T 500 N (500 T)/1.09 ΣM a 0: 6 T (0.3 N N ) 0.78 N T N
63 F B W T 0.3π e W T Hence, W N Eample: Find the force F needed to start the block of mass 100 kg moving to the right if µ s o 50 kg 100 kg F
64 The Square Screw Thread Consider the action of a nut on a screw with square screw thread. z r p α df dn r is the mean radius from the centre-line of the screw thread; p is the pitch (distance along screw between adjacent threads); L is the lead (distance the nut will advance in the direction of ais in one revolution). And for an n-threaded screw, L np
65 Forces are transmitted from screw to nut over several revolutions of the thread. Hence, we have a distribution of normal and friction forces. Due to the narrow width, we may assume that the distribution is confined to the centre-line, thus forming a line load distribution over several turns of the thread. The local slope is given by L tan α π r np π r Infinitesimal normal and friction forces dn and ds have been indicated in figure. All elements of the distributed forces have the same inclination with respect to the z-direction.
66 Hence, the force distribution could be replaced by a single normal force N and a single friction force f at the inclinations shown in figure below. Note also that elements of the distribution have the same moment arm about the centre-line. N P M ΣF z 0: N cos α µ s N sin α P f α ΣM z 0: M z rµ s N cos α + rn sin α
67 These equations are used to eliminate the force N and to get a relation between P and M z. Thus we get M z P r( µ s cosα + sinα ) cosα µ sinα Is the screw self-locking is an important question. That is, having raised the load, if the torque M z is released will the load P stay at the raised position or will it unwind under the action of the load? For this, go back to the equations of equilibrium. Set M z 0, and simultaneously reverse the direction of friction forces. Eliminating N, we get as the condition for impending unwinding of the screw as: s
68 which requires that That is, P r( µ s cosα + sinα) cosα + µ sinα s Thus, if the coefficient of friction µ s equals or eceeds tan α, the screw is of self-locking type; else it will unwind under the load. µ s cos α + sinα µ s tanα Eample: A single threaded jackscrew has a pitch of p mm and a mean diameter of 50 mm. What torque M is needed to lift a load P 5000 N up? If the torque is released, will the load stay in the raised position? µ s
69 P tan α p π r π (5) F Hence, α M z P r( µ s cosα + sinα) cosα µ sinα s (0.5cosα + sinα) cosα 0.5sinα Nmm It is self-locking as µ s > tan α.
STATICS. Friction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.
Eighth E 8 Friction CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University Contents Introduction Laws of Dry Friction.
More informationEngineering Mechanics: Statics
Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines Wedges Used to produce small position adjustments of a body or to apply large forces When sliding is impending, the resultant
More informationEngineering Mechanics. Friction in Action
Engineering Mechanics Friction in Action What is friction? Friction is a retarding force that opposes motion. Friction types: Static friction Kinetic friction Fluid friction Sources of dry friction Dry
More informationChapter 10: Friction A gem cannot be polished without friction, nor an individual perfected without
Chapter 10: Friction 10-1 Chapter 10 Friction A gem cannot be polished without friction, nor an individual perfected without trials. Lucius Annaeus Seneca (4 BC - 65 AD) 10.1 Overview When two bodies are
More informationEngineering Mechanics: Statics
Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines Wedges Used to produce small position adjustments of a body or to apply large forces When sliding is impending, the resultant
More informationM D P L sin x FN L sin C W L sin C fl cos D 0.
789 roblem 9.26 he masses of the ladder and person are 18 kg and 90 kg, respectively. he center of mass of the 4-m ladder is at its midpoint. If D 30, what is the minimum coefficient of static friction
More informationdv dx Slope of the shear diagram = - Value of applied loading dm dx Slope of the moment curve = Shear Force
Beams SFD and BMD Shear and Moment Relationships w dv dx Slope of the shear diagram = - Value of applied loading V dm dx Slope of the moment curve = Shear Force Both equations not applicable at the point
More informationENGINEERING MECHANICS SOLUTIONS UNIT-I
LONG QUESTIONS ENGINEERING MECHANICS SOLUTIONS UNIT-I 1. A roller shown in Figure 1 is mass 150 Kg. What force P is necessary to start the roller over the block A? =90+25 =115 = 90+25.377 = 115.377 = 360-(115+115.377)
More informationOverview. Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance
Friction Chapter 8 Overview Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance Dry Friction Friction is defined as a force of resistance acting on a body which prevents slipping of the body
More informationUnit 21 Couples and Resultants with Couples
Unit 21 Couples and Resultants with Couples Page 21-1 Couples A couple is defined as (21-5) Moment of Couple The coplanar forces F 1 and F 2 make up a couple and the coordinate axes are chosen so that
More informationh p://edugen.wileyplus.com/edugen/courses/crs1404/pc/b02/c2hlch...
If you a empt to slide one... 1 of 1 16-Sep-12 19:29 APPENDIX B If you attempt to slide one solid object across another, the sliding is resisted by interactions between the surfaces of the two objects.
More informationCourse Material Engineering Mechanics. Topic: Friction
Course Material Engineering Mechanics Topic: Friction by Dr.M.Madhavi, Professor, Department of Mechanical Engineering, M.V.S.R.Engineering College, Hyderabad. Contents PART I : Introduction to Friction
More informationBeams. Beams are structural members that offer resistance to bending due to applied load
Beams Beams are structural members that offer resistance to bending due to applied load 1 Beams Long prismatic members Non-prismatic sections also possible Each cross-section dimension Length of member
More informationSTATICS. FE Review. Statics, Fourteenth Edition R.C. Hibbeler. Copyright 2016 by Pearson Education, Inc. All rights reserved.
STATICS FE Review 1. Resultants of force systems VECTOR OPERATIONS (Section 2.2) Scalar Multiplication and Division VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE Parallelogram Law: Triangle
More informationSince the block has a tendency to slide down, the frictional force points up the inclined plane. As long as the block is in equilibrium
Friction Whatever we have studied so far, we have always taken the force applied by one surface on an object to be normal to the surface. In doing so, we have been making an approximation i.e., we have
More informationCHARACTERISTICS OF DRY FRICTION & PROBLEMS INVOLVING DRY FRICTION
CHARACTERISTICS OF DRY FRICTION & PROBLEMS INVOLVING DRY FRICTION Today s Objective: Students will be able to: a) Understand the characteristics of dry friction. b) Draw a FBD including friction. c) Solve
More informationSOLUTION 8 1. a+ M B = 0; N A = 0. N A = kn = 16.5 kn. Ans. + c F y = 0; N B = 0
8 1. The mine car and its contents have a total mass of 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and the tracks is m s = 0.4 when the wheels are locked,
More informationKINGS COLLEGE OF ENGINEERING ENGINEERING MECHANICS QUESTION BANK UNIT I - PART-A
KINGS COLLEGE OF ENGINEERING ENGINEERING MECHANICS QUESTION BANK Sub. Code: CE1151 Sub. Name: Engg. Mechanics UNIT I - PART-A Sem / Year II / I 1.Distinguish the following system of forces with a suitable
More informationThree torques act on the shaft. Determine the internal torque at points A, B, C, and D.
... 7. Three torques act on the shaft. Determine the internal torque at points,, C, and D. Given: M 1 M M 3 300 Nm 400 Nm 00 Nm Solution: Section : x = 0; T M 1 M M 3 0 T M 1 M M 3 T 100.00 Nm Section
More informationSOLUTION 8 7. To hold lever: a+ M O = 0; F B (0.15) - 5 = 0; F B = N. Require = N N B = N 0.3. Lever,
8 3. If the coefficient of static friction at is m s = 0.4 and the collar at is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance x so that the bracket can support
More informationVALLIAMMAI ENGINEERING COLLEGE SRM NAGAR, KATTANKULATHUR DEPARTMENT OF MECHANICAL ENGINEERING
VALLIAMMAI ENGINEERING COLLEGE SRM NAGAR, KATTANKULATHUR 603203 DEPARTMENT OF MECHANICAL ENGINEERING BRANCH: MECHANICAL YEAR / SEMESTER: I / II UNIT 1 PART- A 1. State Newton's three laws of motion? 2.
More informationChapter 5 Equilibrium of a Rigid Body Objectives
Chapter 5 Equilibrium of a Rigid Bod Objectives Develop the equations of equilibrium for a rigid bod Concept of the free-bod diagram for a rigid bod Solve rigid-bod equilibrium problems using the equations
More informationCode No: R Set No. 1
Code No: R05010302 Set No. 1 I B.Tech Supplimentary Examinations, February 2008 ENGINEERING MECHANICS ( Common to Mechanical Engineering, Mechatronics, Metallurgy & Material Technology, Production Engineering,
More informationVector Mechanics: Statics
PDHOnline Course G492 (4 PDH) Vector Mechanics: Statics Mark A. Strain, P.E. 2014 PDH Online PDH Center 5272 Meadow Estates Drive Fairfax, VA 22030-6658 Phone & Fax: 703-988-0088 www.pdhonline.org www.pdhcenter.com
More information2014 MECHANICS OF MATERIALS
R10 SET - 1 II. Tech I Semester Regular Examinations, March 2014 MEHNIS OF MTERILS (ivil Engineering) Time: 3 hours Max. Marks: 75 nswer any FIVE Questions ll Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~~~~
More informationQuestions from all units
Questions from all units S.NO 1. 1 UNT NO QUESTON Explain the concept of force and its characteristics. BLOOMS LEVEL LEVEL 2. 2 Explain different types of force systems with examples. Determine the magnitude
More informationPART-A. a. 60 N b. -60 N. c. 30 N d. 120 N. b. How you can get direction of Resultant R when number of forces acting on a particle in plane.
V.S.. ENGINEERING OLLEGE, KRUR EPRTMENT OF MEHNIL ENGINEERING EMI YER: 2009-2010 (EVEN SEMESTER) ENGINEERING MEHNIS (MEH II SEM) QUESTION NK UNIT I PRT- EM QUESTION NK 1. efine Mechanics 2. What is meant
More information5.2 Rigid Bodies and Two-Dimensional Force Systems
5.2 Rigid odies and Two-Dimensional Force Systems 5.2 Rigid odies and Two-Dimensional Force Systems Procedures and Strategies, page 1 of 1 Procedures and Strategies for Solving Problems Involving Equilibrium
More information4.0 m s 2. 2 A submarine descends vertically at constant velocity. The three forces acting on the submarine are viscous drag, upthrust and weight.
1 1 wooden block of mass 0.60 kg is on a rough horizontal surface. force of 12 N is applied to the block and it accelerates at 4.0 m s 2. wooden block 4.0 m s 2 12 N hat is the magnitude of the frictional
More informationSECTION A. 8 kn/m. C 3 m 3m
SECTION Question 1 150 m 40 kn 5 kn 8 kn/m C 3 m 3m D 50 ll dimensions in mm 15 15 Figure Q1(a) Figure Q1(b) The horizontal beam CD shown in Figure Q1(a) has a uniform cross-section as shown in Figure
More informationif the initial displacement and velocities are zero each. [ ] PART-B
Set No - 1 I. Tech II Semester Regular Examinations ugust - 2014 ENGINEERING MECHNICS (Common to ECE, EEE, EIE, io-tech, E Com.E, gri. E) Time: 3 hours Max. Marks: 70 Question Paper Consists of Part- and
More informationWhen a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.
When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero. 0 0 0 0 k M j M i M M k R j R i R F R z y x z y x Forces and moments acting on a rigid body could be
More informationWhen a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.
When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero. 0 0 0 0 k M j M i M M k R j R i R F R z y x z y x Forces and moments acting on a rigid body could be
More informationWhere, m = slope of line = constant c = Intercept on y axis = effort required to start the machine
(ISO/IEC - 700-005 Certified) Model Answer: Summer 07 Code: 70 Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.
More informationChapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting
More informationOutline: Types of Friction Dry Friction Static vs. Kinetic Angles Applications of Friction. ENGR 1205 Appendix B
Outline: Types of Friction Dry Friction Static vs. Kinetic Angles Applications of Friction ENGR 1205 Appendix B 1 Contacting surfaces typically support normal and tangential forces Friction is a tangential
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5
1 / 40 CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa 2 / 40 EQUATIONS OF MOTION:RECTANGULAR COORDINATES
More informationSN QUESTION YEAR MARK 1. State and prove the relationship between shearing stress and rate of change of bending moment at a section in a loaded beam.
ALPHA COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING MECHANICS OF SOLIDS (21000) ASSIGNMENT 1 SIMPLE STRESSES AND STRAINS SN QUESTION YEAR MARK 1 State and prove the relationship
More informationMini Exam # 1. You get them back in the the recitation section for which you are officially enrolled.
Mini Exam # 1 You get them back in the the recitation section for which you are officially enrolled. One third of you did very well ( 18 points out of 20). The average was 13.4. If you stay in average,
More informationModel Answers Attempt any TEN of the following :
(ISO/IEC - 70-005 Certified) Model Answer: Winter 7 Sub. Code: 17 Important Instructions to Examiners: 1) The answers should be examined by key words and not as word-to-word as given in the model answer
More informationSKAA 1213 Engineering Mechanics
SKAA 1213 Engineering Mechanics TOPIC 6 FRICTION Lecturers: Rosli Anang Dr. Mohd Yunus Ishak Dr. Tan Cher Siang Lesson 7 Outline Introduction Equilibrium on a horizontal plane Equilibrium i on an inclined
More informationUNIT 3 Friction and Belt Drives 06ME54. Structure
UNIT 3 Friction and Belt Drives 06ME54 Structure Definitions Types of Friction Laws of friction Friction in Pivot and Collar Bearings Belt Drives Flat Belt Drives Ratio of Belt Tensions Centrifugal Tension
More informationDry Friction Static vs. Kinetic Angles
Outline: Types of Friction Dry Friction Static vs. Kinetic Angles Applications of Friction 1 Contacting surfaces typically support normal and tangential forces Friction is a tangential force Friction occurs
More informationB.Tech. Civil (Construction Management) / B.Tech. Civil (Water Resources Engineering)
I B.Tech. Civil (Construction Management) / B.Tech. Civil (Water Resources Engineering) Term-End Examination 00 December, 2009 Co : ENGINEERING MECHANICS CD Time : 3 hours Maximum Marks : 70 Note : Attempt
More informationAPPLIED MECHANICS I Resultant of Concurrent Forces Consider a body acted upon by co-planar forces as shown in Fig 1.1(a).
PPLIED MECHNICS I 1. Introduction to Mechanics Mechanics is a science that describes and predicts the conditions of rest or motion of bodies under the action of forces. It is divided into three parts 1.
More informationPhysics 211 Week 10. Statics: Walking the Plank (Solution)
Statics: Walking the Plank (Solution) A uniform horizontal beam 8 m long is attached by a frictionless pivot to a wall. A cable making an angle of 37 o, attached to the beam 5 m from the pivot point, supports
More informationwhere G is called the universal gravitational constant.
UNIT-I BASICS & STATICS OF PARTICLES 1. What are the different laws of mechanics? First law: A body does not change its state of motion unless acted upon by a force or Every object in a state of uniform
More information7.6 Journal Bearings
7.6 Journal Bearings 7.6 Journal Bearings Procedures and Strategies, page 1 of 2 Procedures and Strategies for Solving Problems Involving Frictional Forces on Journal Bearings For problems involving a
More informationb) Fluid friction: occurs when adjacent layers in a fluid are moving at different velocities.
Ch.6 Friction Types of friction a) Dry friction: occurs when non smooth (non ideal) surfaces of two solids are in contact under a condition of sliding or a tendency to slide. (also called Coulomb friction)
More informationConsider the case of a 100 N. mass on a horizontal surface as shown below:
1.9.1 Introduction The study of friction is called: The force of friction is defined as: The force of friction acting between two surfaces has three properties: i) ii) iii) Consider the case of a 100 N.
More informationFIXED BEAMS IN BENDING
FIXED BEAMS IN BENDING INTRODUCTION Fixed or built-in beams are commonly used in building construction because they possess high rigidity in comparison to simply supported beams. When a simply supported
More informationAngular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion
Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for
More informationNewton s Laws.
Newton s Laws http://mathsforeurope.digibel.be/images Forces and Equilibrium If the net force on a body is zero, it is in equilibrium. dynamic equilibrium: moving relative to us static equilibrium: appears
More informationStatic Equilibrium; Torque
Static Equilibrium; Torque The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium. The first condition for equilibrium is that the net force
More informationFE Sta'cs Review. Torch Ellio0 (801) MCE room 2016 (through 2000B door)
FE Sta'cs Review h0p://www.coe.utah.edu/current- undergrad/fee.php Scroll down to: Sta'cs Review - Slides Torch Ellio0 ellio0@eng.utah.edu (801) 587-9016 MCE room 2016 (through 2000B door) Posi'on and
More informationApplications of Calculus I
Applications of Calculus I Application of Maximum and Minimum Values and Optimization to Engineering Problems by Dr. Manoj Chopra, P.E. Outline Review of Maximum and Minimum Values in Calculus Review of
More informationSTATICS. Bodies. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Design of a support
4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies 2010 The McGraw-Hill Companies,
More information5 Equilibrium of a Rigid Body Chapter Objectives
5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body Solve rigid-body equilibrium problems using the
More information2008 FXA THREE FORCES IN EQUILIBRIUM 1. Candidates should be able to : TRIANGLE OF FORCES RULE
THREE ORCES IN EQUILIBRIUM 1 Candidates should be able to : TRIANGLE O ORCES RULE Draw and use a triangle of forces to represent the equilibrium of three forces acting at a point in an object. State that
More informationPhysics, Chapter 3: The Equilibrium of a Particle
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 3: The Equilibrium of a Particle
More informationCreated by T. Madas WORK & ENERGY. Created by T. Madas
WORK & ENERGY Question (**) A B 0m 30 The figure above shows a particle sliding down a rough plane inclined at an angle of 30 to the horizontal. The box is released from rest at the point A and passes
More informationTEST REPORT. Question file: P Copyright:
Date: February-12-16 Time: 2:00:28 PM TEST REPORT Question file: P12-2006 Copyright: Test Date: 21/10/2010 Test Name: EquilibriumPractice Test Form: 0 Test Version: 0 Test Points: 138.00 Test File: EquilibriumPractice
More information24/06/13 Forces ( F.Robilliard) 1
R Fr F W 24/06/13 Forces ( F.Robilliard) 1 Mass: So far, in our studies of mechanics, we have considered the motion of idealised particles moving geometrically through space. Why a particular particle
More information2. Mass, Force and Acceleration
. Mass, Force and Acceleration [This material relates predominantly to modules ELP034, ELP035].1 ewton s first law of motion. ewton s second law of motion.3 ewton s third law of motion.4 Friction.5 Circular
More informationMARKS DISTRIBUTION AS PER CHAPTER (QUESTION ASKED IN GTU EXAM) Name Of Chapter. Applications of. Friction. Centroid & Moment.
Introduction Fundamentals of statics Applications of fundamentals of statics Friction Centroid & Moment of inertia Simple Stresses & Strain Stresses in Beam Torsion Principle Stresses DEPARTMENT OF CIVIL
More information5. Plane Kinetics of Rigid Bodies
5. Plane Kinetics of Rigid Bodies 5.1 Mass moments of inertia 5.2 General equations of motion 5.3 Translation 5.4 Fixed axis rotation 5.5 General plane motion 5.6 Work and energy relations 5.7 Impulse
More informationName: M1 - Dynamics. Date: Time: Total marks available: Total marks achieved:
Name: M1 - Dynamics Date: Time: Total marks available: Total marks achieved: Questions Q1. A railway truck P, of mass m kg, is moving along a straight horizontal track with speed 15 ms 1. Truck P collides
More informationDetermine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
More informationFRICTIONAL FORCES. Direction of frictional forces... (not always obvious)... CHAPTER 5 APPLICATIONS OF NEWTON S LAWS
RICTIONAL ORCES CHAPTER 5 APPLICATIONS O NEWTON S LAWS rictional forces Static friction Kinetic friction Centripetal force Centripetal acceleration Loop-the-loop Drag force Terminal velocity Direction
More informationAP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force).
AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force). 1981M1. A block of mass m, acted on by a force of magnitude F directed horizontally to the
More information1. Replace the given system of forces acting on a body as shown in figure 1 by a single force and couple acting at the point A.
Code No: Z0321 / R07 Set No. 1 I B.Tech - Regular Examinations, June 2009 CLASSICAL MECHANICS ( Common to Mechanical Engineering, Chemical Engineering, Mechatronics, Production Engineering and Automobile
More informationHowever, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied.
FRICTION 1 Introduction In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent
More informationd. Determine the power output of the boy required to sustain this velocity.
AP Physics C Dynamics Free Response Problems 1. A 45 kg boy stands on 30 kg platform suspended by a rope passing over a stationary pulley that is free to rotate. The other end of the rope is held by the
More informationAnnouncements. Principle of Work and Energy - Sections Engr222 Spring 2004 Chapter Test Wednesday
Announcements Test Wednesday Closed book 3 page sheet sheet (on web) Calculator Chap 12.6-10, 13.1-6 Principle of Work and Energy - Sections 14.1-3 Today s Objectives: Students will be able to: a) Calculate
More informationChapter 7 FORCES IN BEAMS AND CABLES
hapter 7 FORES IN BEAMS AN ABLES onsider a straight two-force member AB subjected at A and B to equal and opposite forces F and -F directed along AB. utting the member AB at and drawing the free-body B
More informationIsaac Newton ( )
Isaac Newton (1642-1727) In the beginning of 1665 I found the rule for reducing any degree of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions
More informationENGINEERING MECHANICS - Question Bank
E Semester-_IST YEAR (CIVIL, MECH, AUTO, CHEM, RUER, PLASTIC, ENV,TT,AERO) ENGINEERING MECHANICS - Question ank All questions carry equal marks(10 marks) Q.1 Define space,time matter and force, scalar
More informationExam 3 PREP Chapters 6, 7, 8
PHY241 - General Physics I Dr. Carlson, Fall 2013 Prep Exam 3 PREP Chapters 6, 7, 8 Name TRUE/FALSE. Write 'T' if the statement is true and 'F' if the statement is false. 1) Astronauts in orbiting satellites
More informationLecture 8: Friction Part 1 Friction Phenomena
Lecture 8: Friction Part 1 Friction Phenomena Friction is a force which is generated between. Its direction is always to that of the motion, or tendency for a motion. F mg P N Figure 1 Figure 1 shows a
More informationC7047. PART A Answer all questions, each carries 5 marks.
7047 Reg No.: Total Pages: 3 Name: Max. Marks: 100 PJ DUL KLM TEHNOLOGIL UNIVERSITY FIRST SEMESTER.TEH DEGREE EXMINTION, DEEMER 2017 ourse ode: E100 ourse Name: ENGINEERING MEHNIS PRT nswer all questions,
More informationHATZIC SECONDARY SCHOOL
HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT STATIC EQUILIBRIUM MULTIPLE CHOICE / 33 OPEN ENDED / 80 TOTAL / 113 NAME: 1. State the condition for translational equilibrium. A. ΣF = 0 B. ΣF
More information2016 ENGINEERING MECHANICS
Set No 1 I B. Tech I Semester Regular Examinations, Dec 2016 ENGINEERING MECHANICS (Com. to AE, AME, BOT, CHEM, CE, EEE, ME, MTE, MM, PCE, PE) Time: 3 hours Max. Marks: 70 Question Paper Consists of Part-A
More informationUNIT 2 FRICTION 2.1 INTRODUCTION. Structure. 2.1 Introduction
UNIT FICTION Structure.1 Introduction Objectives. Types of.3 Laws of Dry.4 Static and Kinetic.5 Coefficient of.6 Angle of epose.7 Least Force equired to Drag a Body on a ough Horizontal Plane.8 Horizontal
More informationChapter 2. Shear Force and Bending Moment. After successfully completing this chapter the students should be able to:
Chapter Shear Force and Bending Moment This chapter begins with a discussion of beam types. It is also important for students to know and understand the reaction from the types of supports holding the
More informationSimple Machines. Changes effort, displacement or direction and magnitude of a load 6 simple machines. Mechanical Advantage
Simple Machine Simple Machines Changes effort, displacement or direction and magnitude of a load 6 simple machines Lever Incline plane Wedge Screw Pulley Wheel and Axle Mechanical Advantage Ideal: IMA
More informationWelcome back to Physics 211
Welcome back to Physics 211 Today s agenda: Weight Friction Tension 07-1 1 Current assignments Thursday prelecture assignment. HW#7 due this Friday at 5 pm. 07-1 2 Summary To solve problems in mechanics,
More informationPERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK
PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM - 613 403 - THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub : Strength of Materials Year / Sem: II / III Sub Code : MEB 310
More informationBeams are bars of material that support. Beams are common structural members. Beams can support both concentrated and distributed loads
Outline: Review External Effects on Beams Beams Internal Effects Sign Convention Shear Force and Bending Moment Diagrams (text method) Relationships between Loading, Shear Force and Bending Moments (faster
More informationCHAPTER 4 NEWTON S LAWS OF MOTION
62 CHAPTER 4 NEWTON S LAWS O MOTION CHAPTER 4 NEWTON S LAWS O MOTION 63 Up to now we have described the motion of particles using quantities like displacement, velocity and acceleration. These quantities
More informationChapter - 1. Equilibrium of a Rigid Body
Chapter - 1 Equilibrium of a Rigid Body Dr. Rajesh Sathiyamoorthy Department of Civil Engineering, IIT Kanpur hsrajesh@iitk.ac.in; http://home.iitk.ac.in/~hsrajesh/ Condition for Rigid-Body Equilibrium
More informationIshik University / Sulaimani Architecture Department. Structure. ARCH 214 Chapter -5- Equilibrium of a Rigid Body
Ishik University / Sulaimani Architecture Department 1 Structure ARCH 214 Chapter -5- Equilibrium of a Rigid Body CHAPTER OBJECTIVES To develop the equations of equilibrium for a rigid body. To introduce
More informationHSC PHYSICS ONLINE B F BA. repulsion between two negatively charged objects. attraction between a negative charge and a positive charge
HSC PHYSICS ONLINE DYNAMICS TYPES O ORCES Electrostatic force (force mediated by a field - long range: action at a distance) the attractive or repulsion between two stationary charged objects. AB A B BA
More informationStatics deal with the condition of equilibrium of bodies acted upon by forces.
Mechanics It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics
More information3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM
3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity Two types of forces that act on it, the
More informationJNTU World. Subject Code: R13110/R13
Set No - 1 I B. Tech I Semester Regular Examinations Feb./Mar. - 2014 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Max. Marks: 70 Question
More informationPLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY Today s Objectives: Students will be able to: 1. Define the various ways a force and couple do work.
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY Today s Objectives: Students will be able to: 1. Define the various ways a force and couple do work. In-Class Activities: 2. Apply the principle of work
More information1. Draw a FBD of the toy plane if it is suspended from a string while you hold the string and move across the room at a constant velocity.
1. Draw a FBD of the toy plane if it is suspended from a string while you hold the string and move across the room at a constant velocity. 2. A 15 kg bag of bananas hangs from a taunt line strung between
More informationTypes of Structures & Loads
Structure Analysis I Chapter 4 1 Types of Structures & Loads 1Chapter Chapter 4 Internal lloading Developed in Structural Members Internal loading at a specified Point In General The loading for coplanar
More informationSECOND ENGINEER REG. III/2 APPLIED MECHANICS
SECOND ENGINEER REG. III/2 APPLIED MECHANICS LIST OF TOPICS Static s Friction Kinematics Dynamics Machines Strength of Materials Hydrostatics Hydrodynamics A STATICS 1 Solves problems involving forces
More informationCE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR
CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR 2014-2015 UNIT - 1 STRESS, STRAIN AND DEFORMATION OF SOLIDS PART- A 1. Define tensile stress and tensile strain. The stress induced
More information